1、整数划分问题
#define _CRT_SECURE_NO_WARNINGS
#include
int p(int n, int m)
{
if (m == 1 || n == 1)
return 1;
if (m > n)
return p(n, n);
if (m == n)
return 1 + p(n, m - 1);
return p(n, m - 1) + p(n - m, m);
}
int main(void)
{
int n, s;
scanf("%d", &n);
s = p(n, n);
printf("the number of division of %d is %d\n",n,s);
getchar();
}
2、递归折半查找
#define _CRT_SECURE_NO_WARNINGS
#include
int bin_search(int key[], int low, int high, int k)
{
int mid;
if (low > high)
return -1;
else
{
mid = (low + high) / 2;
if (key[mid] == k)
return mid;
if (k > key[mid])
return bin_search(key, mid + 1, high, k);
else
return bin_search(key, low, mid - 1, k);
}
}
int main(void)
{
int n, i, addr;
int A[10] = { 5,6,7,8,15,19,28,98,104,109 };
printf("the contents of the Array A[10] are\n");
for (i = 0; i < 10; i++)
{
printf("%d\n", A[i]);
}
printf("please input:\n");
scanf("%d", &n);
addr = bin_search(A, 0, 9, n);
if (-1 != addr)
printf("%d is at the %dth unit in array A\n", n, addr+1);
else
printf("error!");
getchar();
}
3、查找指定范围素数
#define _CRT_SECURE_NO_WARNINGS
#include
//判断n是否是素数
int isPrime(int n)
{
int i;
for (i = 2; i < n; i++)
{
if (n%i == 0)
return 0;
}
return 1;
}
//寻找[low,high]之间的素数
getPrime(int low, int high)
{
int i;
for (i = low; i <= high; i++)
if (isPrime(i))
printf("%d\n", i);
}
int main(void)
{
int low, high;
printf("please input the domain for searching prime\n");
printf("low limitation:");
scanf("%d", &low);
printf("high limitation:");
scanf("%d", &high);
printf("The whole primes in this domain are\n");
getPrime(low, high);
getche();
}
4、简化版桶排序
4.1 考试成绩十分制 输入成绩 从小到大排序
#include
int main(void)
//此算法的核心思想是元素序号即代表实际对应值，
//元素储存的值为实际出现次数
{
int a[11], i, j, t;
for (i = 0; i <= 10; i++)
{
a[i] = 0;//数组每个元素初始化为0
}
//循环读入10个数
for (i = 1; i <= 10; i++)
{
scanf("%d", &t);
a[t]++;//进行计数
}
for (i = 0; i < 11; i++)//依次判断a[0]~a[10]
{
for (j = 1; j <= a[i]; j++)
printf("%d\n", i);
}
getchar();
return 0;
}
改进后可以输入自定义个数的成绩
#include
int main(void)
//此算法的核心思想是元素序号即代表实际对应值，
//元素储存的值为实际出现次数
{
int a[11], i, j, t,n;
for (i = 0; i <= 10; i++)
{
a[i] = 0;//数组每个元素初始化为0
}
scanf("%d", &n);
//循环读入n个数
for (i = 1; i <= n; i++)
{
scanf("%d", &t);
a[t]++;//进行计数
}
for (i = 0; i < 11; i++)//依次判断a[0]~a[10]
{
for (j = 1; j <= a[i]; j++)
printf("%d\n", i);
}
getchar();
return 0;
}
- 上述桶排序的时间复杂度为O(M+N) M为桶的个数 N为待排序的数的个数
5、指针实现冒泡排序（从小到大）
#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
void bubble(int *p, int len)
{
for (int i = 0; i < len - 1; i++)
{
for (int j = 0; j < len - 1 - i; j++)
{
if (*(p + j) > *(p + j + 1))
{
int temp = *(p + j);
*(p + j) = *(p + j + 1);
*(p + j + 1) = temp;
}
}
}
}
int main(void)
{
int arr[] = { 3,14,1,49,20,0,18,22,98,30 };
int *p = arr;
int len = sizeof(arr) / sizeof(int);
bubble(p, len);
for (int i = 0; i < 10; i++)
{
printf("%d\n", *(p+i));
}
system("pause");
return EXIT_SUCCESS;
}
6.矩阵连乘问题
#include
#include
#include
using namespace std;
const int msize = 100;
int p[msize];
int m[msize][msize], s[msize][msize];
int n;
void matrixchain()
{
int i, j, r, k;
memset(m, 0, sizeof(m));
memset(s, 0, sizeof(s));
for (r = 2; r <= n; r++) //不同规模的子问题
{
for (i = 1; i <= n - r + 1; i++)
{
j = i + r - 1;
m[i][j] = m[i + 1][j] + p[i - 1] * p[i] * p[j]; //决策为 k=i 的乘法次数
s[i][j] = i; //子问题的最优策略是 i;
for (k = i + 1; k < j; k++) //对从 i 到 j 的所有决策，求最优值，记录最优策略
{
int t = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
if (t < m[i][j])
{
m[i][j] = t;
s[i][j] = k;
}
}
}
}
}
void print(int i, int j)
{
if (i == j)
{
cout << "A[" << i << "]";
return;
}
cout << "(";
print(i, s[i][j]);
print(s[i][j] + 1, j);
cout << ")";
}
int main()
{
cout << "请输入矩阵的个数 n：";
cin >> n;
int i, j;
cout << "请依次输入每个矩阵的行数和最后一个矩阵的列数：";
for (i = 0; i <= n; i++)
cin >> p[i];
matrixchain();
print(1, n);
cout << endl;
cout << "最小计算量的值为：" << m[1][n] << endl;
}
7.电路布线问题
#include
#include
#include
using namespace std;
void mnset(int C[], int n, int* size)
{
int i, j;
// 当i = 1时
for (j = 0; j < C[1]; j++)
{
size[n + j] = 0;
}
for (j = C[1]; j <= n; j++)
{
size[n + j] = 1;
}
// 当i >=2时
for (i = 2; i < n; i++)
{
for (j = 0; j < C[i]; j++)
{
size[i * n + j] = size[(i - 1) * n + j];
}
for (j = C[i]; j <= n; j++)
{
size[i * n + j] = max(size[(i - 1) * n + j], size[(i - 1) * n + C[i] - 1] + 1);
}
}
size[n * n] = max(size[(n - 1) * n], size[(n - 1) * n + C[n - 1] - 1] + 1);
}
void traceBack(int C[], int* size, int n, int Net[], int& m)
{
int i, j = n;
m = 0;
for (i = n; i > 1; i--)
{
if (size[i * (n + 1) + j] != size[(i - 1) * (n + 1) + j])
{
Net[m++] = i;
j = C[i] - 1;
}
}
if (j >= C[1])
{
Net[m++] = 1;
}
}
int main()
{
int n; // 接线数目
int num; // 最大子集数
int* l; // 接线方式
int* p; // 子集数
int* size; // 二维表数组
cout << "请输入电路的接线柱数:";
cin >> n;
// 动态创建热线方式数组
l = new int[n + 1];
// 动态创建子集数组
p = new int[n + 1];
// 临时动态创建二维数组
int** temp;
temp = new int* [n + 1];
for (int i = 0; i < n + 1; i++)
{
temp[i] = new int[i + 1];
}
// 将二维表二维数组以一维指针赋给size
size = &temp[0][0];
cout << "请依次输入连接数：" << endl;
for (int i = 1; i <= n; i++)
cin >> l[i];
// 调用计算最优值函数
mnset(l, n + 1, size);
// 调用构造最优值函数
traceBack(l, size, n, p, num);
// 打印子集数
cout << "最大不想交子集数：" << num << endl << endl;
//打印出子集
cout << "最大不想交子集" << endl;
for (int i = 0; i < num; i++)
cout << p[i] << " --> " << l[p[i]] << endl;
system("pause");
return 0;
}
8.最长公共子序列问题
#include
#include
#include
using namespace std;
const int N = 1002;
int c[N][N];
int b[N][N];
char s1[N], s2[N];
int len1, len2;
void LCSL()
{
int i, j;
for (i = 1; i <= len1; i++)//控制 s1 序列
for (j = 1; j <= len2; j++)//控制 s2 序列
{
if (s1[i - 1] == s2[j - 1])
{//如果当前字符相同，则公共子序列的长度为该字符前的最长公共序列+1
c[i][j] = c[i - 1][j - 1] + 1;
b[i][j] = 1;
}
else
{
if (c[i][j - 1] >= c[i - 1][j])
{
c[i][j] = c[i][j - 1];
b[i][j] = 2;
}
else
{
c[i][j] = c[i - 1][j];
b[i][j] = 3;
}
}
}
}
void print(int i, int j)//根据记录下来的信息构造最长公共子序列（从 b[i][j]开始递推）
{
if (i == 0 || j == 0) return;
if (b[i][j] == 1)
{
print(i - 1, j - 1);
cout << s1[i - 1];
}
else if (b[i][j] == 2)
print(i, j - 1);
else
print(i - 1, j);
}
int main()
{
int i, j;
cout << "输入字符串 s1：" << endl;
cin >> s1;
cout << "输入字符串 s2：" << endl;
cin >> s2;
len1 = strlen(s1);//计算两个字符串的长度
len2 = strlen(s2);
for (i = 0; i <= len1; i++)
{
c[i][0] = 0;//初始化第一列为 0
}
for (j = 0; j <= len2; j++)
{
c[0][j] = 0;//初始化第一行为 0
}
LCSL(); //求解最长公共子序列
cout << "s1 和 s2 的最长公共子序列长度是：" << c[len1][len2] << endl;
cout << "s1 和 s2 的最长公共子序列是：";
print(len1, len2); //递归构造最长公共子序列最优解
return 0;
}
9.快速排序
def quick_sort(array):
if len(array) < 2:
return array
else:
pivot = array[0]
less = [i for i in array[1:] if i <= pivot]
greater = [i for i in array[1:] if i >= pivot]
return quick_sort(less) + [pivot] + quick_sort(greater)